Consider a hydrogen like atom whose energy in n^ th exicited stat… - Vidyadip

MCQ

Consider a hydrogen like atom whose energy in $n^{th}$ exicited state is given by ${E_n} = - \frac{{13.6{Z^2}}}{{{n^2}}}$ when this excited atom makes a transition from excited state to ground state, most energetic photons have energy $E_{max} = 52.224\, eV$ and least energetic photons have energy $E_{min} = 1.224\, eV$. The atomic number of atom is

✓
$2$
B
$5$
C
$4$
D
None of these

✓

Answer

Correct option: A.

$2$

a
(a) Maximum energy is liberated for transition ${E_n} \to 1$ and minimum energy for ${E_n} \to {E_{n - 1}}$

Hence $\frac{{{E_1}}}{{{n^2}}} - {E_1} = 52.224\,eV$……$(i)$

and $\frac{{{E_1}}}{{{n^2}}} - \frac{{{E_1}}}{{{{(n - 1)}^2}}} = 1.224\,eV$…..$(ii)$

Solving equations $(i)$ and $(ii)$ we get

${E_1} = - 54.4\,eV$ and $n = 5$

Now ${E_1} = - \frac{{13.6\,{Z^2}}}{{{1^2}}} = - 54.4\,eV.$

Hence $Z = 2$

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